Oracle 1Z0-882 Online Practice Questions and Exam Preparation

Oracle 1Z0-882 Online Questions & Answers

• Question 1:

  When executing a stored routine, how is the SQL_MODE determined?

  A. By the default SQL_MODE of the server
  B. By the current SQL _MODE of the session
  C. By the SQL_MODE that was set when the routine was defined
  D. By using TRADITTIONAL regardless of any other settings
  B. By the current SQL _MODE of the session

• Question 2:

  Which three are valid identifiers for the user table in the mysq1 database?

  A. myssq1. user
  B. `mysq1. user'
  C. `mysq1'. `user'
  D. Mysq1. `user'
  E. `'mysq1. User''
  A. myssq1. user
  D. Mysq1. `user'

• Question 3:

  Assume the user has just connected to the MySQL server. What is the result of the query SELECT @ a?

  A. An error that @ a is undefined
  B. A single NULL
  C. An empty string
  D. The value of GLOBAL variable @ a
  D. The value of GLOBAL variable @ a

• Question 4:

  You want to query the VARCHAR column ` code' values that match:

  

  Start with "p"

  

  End with "_"

  

  Contain more than 3 characters

  Assume that sql_mode is blank.

  Which two queries select only those rows?

  A. SELECT code FROM operations WHERE code LIKE "pandand_";
  B. SELECT code FROM operations WHERE code LIKE " 'and'and_' "ESCAPE " ` '';
  C. SELECT code FROM operations WHERE code LIKE "pand_\_";
  D. SELECT code FROM operations WHERE code LIKE "p_\and\_";
  E. SELECT code FROM operations WHERE code LIKE "p_7_and;_"ESCAPE "/";
  C. SELECT code FROM operations WHERE code LIKE "pand_\_";
  E. SELECT code FROM operations WHERE code LIKE "p_7_and;_"ESCAPE "/";

• Question 5:

  Consider the content of the class and student tables: Class Which three queries produce the same result?

  

  A. SELECT * FROM class INNER JOIN student ON class.class_id=student.class_id
  B. SELECT * FROM JOIN student LEFT JOIN student ON class. Class.class_id=student.class_id
  C. SELECT * FROM class INNER JOIN student WHERE NOT ISNULL (student.class_id)
  D. SELECT * FROM JOIN student On class .class_id=student.class_id WHERE NOT ISNULL (student.class_id)
  E. SELECT * FROM student RIGHT JOIN class ON class.class_id=student.class_id
  D. SELECT * FROM JOIN student On class .class_id=student.class_id WHERE NOT ISNULL (student.class_id)

• Question 6:

  An application tracks usage of educational courses in a company. Many people can take one course. Each person can take multiple courses. The data has been stored in one table but it is growing too large. You decide to normalize the table.

  What would a normalized data model contain?

  A. Two tables: employee and course, with foreign keys on employee ID and course ID.
  B. Three tables: employee, course, and a table cross-referencing employee IDs and course IDs
  C. Two tables: an employee table with multiple course IDs , and a course table
  D. Four tables: employee, course, courses by employee, and employees by course
  A. Two tables: employee and course, with foreign keys on employee ID and course ID.

• Question 7:

  Cormcctor/Net supports a decoupled database connection mode, where a database Connection is established only when needed. Choose the three parts that are involved when working with decoupled data.

  A. OpenConnect.MySql, DataTTiread.MySql, CommandExecutor.MySql
  B. MySqlQueryBuIld, MySqlQuerytxer. MySq I Query Results
  C. DecoupODBCConn, DecoupMyScjlruiiimanrt, DecoupMySqlRetum
  D. DataSet, MySqlDataAdapter, MySql Command Builder
  D. DataSet, MySqlDataAdapter, MySql Command Builder

• Question 8:

  You wish to create a trigger on the `city' table that will check the value of the `District' field before any INSERT. The trigger needs to change it to" Unknown" for an empty string or NULL.

  CREATE TRIGGER City_bi BEFORE INSERT ON CITY FOR EACH ROW BEGIN IF OLD. District IS NULL OR OLD.District= . . THEN SET NEW.District='Unknown'; END IF : END;

  Does the CREATE TRIGGER statement accomplish this goal?

  A. Yes; the trigger works correctly.
  B. No; FOR EACH ROW is invalid syntax.
  C. No; the syntax should be CREATE TRIGGER city-bi ON city BEFORE INSERT....
  D. No; the OLD keyword cannot be used in an INSERT trigger.
  A. Yes; the trigger works correctly.

• Question 9:

  Examine this table that contains over two million rows of data:

  CREATE TABLE `news_feed' (

  .id'bigint (20) NOT NULL AUTO _INCREMENT,

  .news _sources_id'varchar (11) NOT NULL,

  .dataline' datetime NOT NULL,

  .headline' varchar (256) NOT NULL,

  .story' text NOT NULL,.tag varchar (32768) DEFAULT NULL, PRIMARY KEY (`id') KEY `dateline' ( `dateline')

  )

  Examine this query that returns 332 rows of date:

  SELECT *

  FROM news_feed

  WHERE DATE(dateline)= `2013-01-01'

  Which change would show the greatest improvement in the response time of the query?

  A. Use the LIKE operator: SELECT . . .WHERE dateline LIKE `2013-10-01and'
  B. USE the DATEDIFF function: SELECT . . . WHERE DATEDIFF (dateline, `2013-01-01') = 0
  C. Use numeric equivalents for comparing the two dates: SELECT. . .WHERE MOD(UNIX_TIMESTAMP (dateline), 86400 =UNIX_TIMESTAMP (`2013-01-01')
  D. Use a date range comparison: SELECT . . . WHERE dateline >= `2013-01' and dateline < `2013-01-02'
  D. Use a date range comparison: SELECT . . . WHERE dateline >= `2013-01' and dateline < `2013-01-02'

• Question 10:

  You have two lists of values to correlate.

  

  Which query lists all names in colors1 and how many total matches are there in colors2?

  A. SELECT colors1 .name.count (colors2.name) FROM colors1. Colors2 WHERE Colors1. Name = (SELECT DISTINCT name FROM colors2 WHERE colors1.name=colors2.name) GROUP BY colorse1.name,
  B. SELECT colors1.name, count(colorse2. Name) FROM colorse1 .name =colors2.name WHERE colors1. Name =colors2.name GROUP BY colors1.name,
  C. SELECT colors1. Name count (colors2.name) FROM colors1 INNER JOIN colors2 on colors1. Name =colors2. Name GROUP BY colors1 .name;
  D. SELECT colors1.name, count (colors2.name) FROM JOIN colors2 on colors1 .name =colors2.name GROUP BY colors1.name;
  E. SELECT colors1.name, count (colors2.name) FROM colors1 RIGHT JOIN colors1 on colors1 .name =colors2.name GROUP BY colors1.name;
  D. SELECT colors1.name, count (colors2.name) FROM JOIN colors2 on colors1 .name =colors2.name GROUP BY colors1.name;